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Math Help - Limit Question

  1. #1
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    Limit Question

    I don't understand this problem. Can someone please help me?

    Problem:

    Lim [ cos(π+Δx) +1 ] / Δx
    Δx -> 0

    Hint: cos(x+y)=cosx cosy-sinx siny

    My attempt:

    [cosπ cosΔx + sinπ sinΔx + 1]/ Δx

    [-1 cosΔx + 0 sinΔx +1 ]/ Δx

    (-cosΔx + 1)/ Δx

    I don't know what to do from here. I want to get rid of Δx from the denominator, but I don't know how to. Can someone please help me?
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  2. #2
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    Use l'Hopitals?

    <br />
\lim\limits_{x\to0} \frac{-cos(x)+1}{x}=\lim\limits_{x\to0} \frac{sin(x)}{1} = 0<br />

    Is that what you mean?
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  3. #3
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    I wasn't taught that rule yet, so I doubt that is the way she wanted us to solve the problem. Can you please explain what you did to get that answer though?
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  4. #4
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    Quote Originally Posted by Power of One View Post
    I wasn't taught that rule yet, so I doubt that is the way she wanted us to solve the problem. Can you please explain what you did to get that answer though?
    L'Hospital's Rule states that if a limit is of indeterminate form \frac{0}{0} or \frac{\infty}{\infty}, then

    \lim_{x \to 0}\frac{f(x)}{g(x)} = \lim_{x \to 0}\frac{f'(x)}{g'(x)}.

    So in your case

    \lim_{\Delta x \to 0}\frac{-\cos{\Delta x} + 1}{\Delta x} = \lim_{\Delta x \to 0}\frac{\frac{d}{d\Delta x}(- \cos{\Delta x} + 1)}{\frac{d}{d\Delta x}(\Delta x)}

     = \lim_{\Delta x \to 0}\frac{\sin{\Delta x}}{1}

     = 0


    As it is, you don't need to resort to L'Hospital's Rule, as the original limit can also be solved using the sandwich theorem. Use Google to find some more info.
    Last edited by Prove It; September 21st 2009 at 07:23 PM.
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  5. #5
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    sin(x)/1 goes to 0 as x goes to 0.
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  6. #6
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    So it does. I had the famous limit \lim_{x \to 0}\frac{\sin{x}}{x} in my head at the time...
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